The graph of $(x-3)^2 + (y-5)^2=16$ is reflected over the line $y=2$. The new graph is the graph of the equation $x^2 + Bx + y^2 + Dy + F = 0$ for some constants $B$, $D$, and $F$. Find $B+D+F$.
Answers
The graph of (x-3)^2 + (y-5)^2=16 is reflected over the line y=2. The new graph is the graph of the equat ion x^2 + Bx + y^2 + Dy + F = 0 for some const ants B, D, and F. Find B+D+F.
The original graph is a circle with a cen ter at (3, 5) and a radius of 4 units.
If it is reflected across the line y = 2, the center of the new circle is (3, -1) and the = this we have
x^2 - 6x + 9 + y^2 + 2y + 1 = 16 sim plify
x^2 - 6x + y^2 + 2y + 10 = 16
x^2 - 6x + y^2 + 2y - 6 = 0 and B = -6 and D = 2 and F = -6 so B + D + F = -6 + 2 - 6 = -10
The original circle is centered at (3,5) with a radius of 4.
Reflecting across the y=2 shifts the y coordinate of the center.
The center is 3 units above the line y=2.
The reflected circle will have the center 3 units below the line y=2.
2-3=-1
.
.
.
%28x-3%29%5E2%2B%28y%2B1%29%5E2=16
x%5E2-6x%2B9%2By%5E2%2B2y%2B1=16
highlight%28x%5E2-6x%2By%5E2%2B2y-6=0%29
.
.
.
I leave it to you to determine the sum B%2BD%2BF
.
.
.
✨✨Hope it helps you✨✨
